Question

How do you factor `64x^3-1`?

Answer 1

`64x^3 - 1 = (4x-1)(16x^2+4x+1)`

`= (4x-1)(4x + 1/2 - sqrt(3)/2i)(4x + 1/2 + sqrt(3)/2i)`

Explanation:

The difference of cubes formula states that:
`a^3 - b^3 = (a-b)(a^2+ab+b^2)`
(try multiplying the right side to verify this)

Applying this, we have:

`64x^3 - 1 = (4x)^3 - 1^3`
`= (4x-1)((4x)^2+4x*1+1^2)`
`= (4x-1)(16x^2+4x+1)`

If we wished to factor further, we could apply the quadratic formula to find the roots of `16x^2+4x+1`.

`16x^2 + 4x + 1 = 0 => x = (-4+-sqrt(4^2-4(16)(1)))/(2*16)`

`=> x = (-4+-sqrt(-48))/32`

`=> x = (-1+-sqrt(-3))/8`

As we have the square root of a negative number, we have no real solutions, and so are done if we are using only real numbers. If we allow for complex numbers, then using `i = sqrt(-1)`, we have:

`x = -1/8+-sqrt(3)/8i`

and thus, remembering to multiply by `16` to account for the coefficient of `x^2`, we can completely factor the original expression as

`64x^3 -1 = (4x-1)(4x + 1/2 - sqrt(3)/2i)(4x + 1/2 + sqrt(3)/2i)`